Good question!

Here is probably not the shortest way:

5 colors: \(C^6_5 = 6\) - 5 colors out of 6.
4 colors: \(C^6_4 * C^4_1 = 15*4 = 60\) - 4 colors out of 6 and one of 4 colors are used twice.
3 colors: \(C^6_3 * (C^3_1 + C^3_1) = 20*6 = 120\) - 3 colors out of 6 and one color of 3 for 3+1+1 combination and one color of 3 for 2+2+1 combination.
2 colors: \(C^6_2 * (C^2_1 + C^2_1) = 15*4 = 60\) - 2 colors out of 6 and one color of 2 for 3+2 combination and one color of 2 for 4+1 combination.
1 color: \(C^6_1 = 6\) - 1 color out of 6.

6+60+120+60+6 = 252
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If we have only 3 colors for 5 boards, there are only 2 possible options 3 (first color) + 1 (second color) + 1 (third color) and 2(first color) + 2 (second color) + 1 (third color). There is 3 different ways for each option.

For example, option 1 if we have red, green, blue:

red + green + blue + blue + blue
red + green + green + green + blue
red + red + red + green + blue

option 2:

red + red + green + green + blue
red + red + green + blue + blue
red + green + green + blue + blue

So, \(C^3_1\) actually means that we choose one color out of 3 for 3 boards (option 1) or for 1 board (option 2)
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Yup, \(C^3_2\) is correct too.
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